 # AP Board Class 9 Maths Chapter 2 Polynomials and Factorisation

AP Board Class 9 Maths Chapter 2, discusses the concept of polynomials, its definition and several other aspects like the degree of a polynomial, zeros of a polynomial and the division algorithm for Polynomials. In addition to this, some of the other important topics covered in Chapter 2 include;

• Factorisation of Polynomials.
• Remainder and Factor Theorem.
• Some Algebraic Identities.

## What Is A Polynomial?

An algebraic expression, which contains two or more terms is known as a polynomial. A polynomial is usually made up of variables, constants, exponents and is a result of either addition, subtraction, multiplication and division. There are several types of polynomials like monomials containing one term, binomials containing only two terms and trinomials which contains three terms. Further, there is linear polynomial which is a polynomial of degree 1. Similarly, there is quadratic polynomial and cubic polynomial, which are polynomials with degree 2 and degree 3 respectively.

Talking about polynomials in one variable these are expressions, which consist of only one type of variable. However, polynomials can have zero or more than two variables.

### Factorisation of Polynomials

Factorisation of polynomials is the process of expressing a polynomial or breaking a polynomial into the product of its factors.

• Factor Theorem: If x – a is a factor of the polynomial p(x), then p(a) = 0. Also if p(a) = 0 then (x – a) is a factor of p(x).
• Remainder Theorem: If p(x) is any polynomial of degree greater than or equal to 1 and p(x) is divided by the linear polynomial (x – a), then the remainder is p(a).

### Some Algebraic Identities

 Identity I (a + b)2 = a2 + 2ab + b2 Identity II (a – b)2 = a2– 2ab + b2 Identity III (a + b)(a – b) = a2 – b2 Identity IV (x + a)(x + b) = x2 + (a + b) x + ab

Students can have a look at some of the chapter questions along with their solutions below;

Question 1: Which of these is a monomial, 2x2 or 5x-3?

Solution: 2x2 because the exponent of the variable is a positive integer.

Question 2: Factor 3x3 – x2y +6x2y – 2xy2 + 3xy2 – y3 =

Solution: 3x3 – x2y + 6x2y – 2xy2 + 3xy2 – y3

=x2(3x – y) + 2xy(3x – y) + y2(3x-y)

=(3x – y)(x2 + 2xy + y2)

=(3x – y)(x + y)2

Question 3: Find the remainder when x3 + 1 divided by (x + 1)

Solution:

Here p(x) = x 3 + 1

The zero of the linear polynomial x + 1 is -1 [x + 1 = 0, x = -1]

So replacing x by -1

p(-1) = (-1)3 + 1

= -1 + 1

= 0

Question 4: Expand (2a + 3b + 5)using identity.

Solution: : When we compare this expression with (x + y + z)2, the result given is  x = 2a, y = 3b and z = 5
Hence, using Identity V, we get (2a + 3b + 5)2.  Here, x= (2a), y=  (3b) and z =5.  Thus, applying the identity V, we get
(2a +3b+5) 2 = (2a)2 + (3b)2 + (5)2 + 2(2a)(3b) + 2(3b)(5) + 2(5) (2a)
= 4a2+ 9b2+ 25 + 12ab + 30b + 20a

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